the intensity level of an orchestra is 85 dB. A single violin achieves a level o

Question

the intensity level of an orchestra is 85 dB. A single violin achieves a level of 70 dB. How does the intensity of the sound of the full orchestra compare with that of the violin's sound? I get that we are comparing a ratio, I'm having trouble with the logs. From our notes, B2 =85dB and B1 = 70 dB. B2 = 10 log (I2/I0) and B1 = 10 log (I1/I0) I get that.
B2 -B1 = 10 log (I2/I1) = 15 dB (get that) or log (I2/I1) = 1.5 (what?)
So (I2/I1) = 10 (1.5) = 31.6 (I had to enter 1 log 1.5 to get 31.6, 10 log 1.5 came up with 316) or I orchestra = 321 violin (what?).

I get that we are comparing the ratio and don't have to compute I1/I0. I'm not certain why I had to alter my input into my calculator and I'm missing the final part of the ratio set up. Thanks for the help!

Explanation / Answer

B=(10dB) log(Ia/Ib) Ia = sound intensity Ib = threshold of human hearing (1.0 * 10^-12 W/m^2) Calculating Ia of orchestra we get 85 = 10 log (Iao/10^-12) 8.5=log(Iao/10^-12) Iao = 3.162 x 10^-4 Calculating Iav of violon we get 70 = 10 log (Iav/10^-12) 7=log(Iav/10^-12) 10^7=Iav/10^-12 Iav = 10^-5. Comparing intensity of thesound of full orchestra with that of the violins sound we get Iao/Iav = (3.162 x 10^-4)/ (10^-5) = 31.62

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